Consider two antennas separated by a distance of 200 m which are each generating radio waves with a frequency of 6.9 MHz in phase with each other. The two antennas are located with one directly west of the other one. What are the two farthest points directly north of one of the antenna where there will be destructive interference, so that a radio receiver placed at that point would not pick up any signal? What is the farthest and next farthest Y? Give anser in meters. I just need a direction to go with this question, I'm stumped.

Question

anonymous · Answer

Firstly, the question is concerned with phase differences between the waves from the two antennae due to differences in path length from antenna to receiver.
This is given by kd, where k is 2pi/wavelength and d is the path difference.

Secondly, in order to obtain destructive interference, we need that phase difference to be an odd multiple of pi.

Thirdly, if you consider a point that is due north of one of the antennae, you need to appreciate that as you move further and further away, the path difference will decrease. If you are a very very long way from the antennae, the paths will be of almost the same length - and so as you move away from the antennae there comes a point where you cannot get sufficient path difference to give any destructive interference. You are asked to find the furthest location where destructive interference is still possible.

anonymous · Answer

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The path difference is r1-r2

anonymous · Answer

A1 and A2 are the antennae, and R is the position of the receiver.

anonymous · Answer

Still stumped ?

radar · Answer

For complete destructive interference, would not the amplitude be required to be equal for each wave as well as exactly out of phase?

radar · Answer

If that is the case then the path of each wave must also attenuate each signal equally, this makes this ptoblem kind of difficult don't it.

radar · Answer

The point receiver must not pick up any signal so says the problem ??

radar · Answer

Well considering one path would only have to differ by 1/2 wavelength from the other path, the distance difference for each wave would allow the amplitudes to be close enough in value to cancel.  So please ignore what I am saying, but go ahead and calculate the wavelength of a 6.9 MHz signal.

anonymous · Answer

@radar Yes, you're right that the amplitudes should be equal for complete cancellation, but as you say, I assumed that the difference in amplitudes would be small enough to ignore and focused my hints on the path difference. : )

radar · Answer

@ProfBrainstorm  I agree the power was not to be considered, for one, they did not mention that the power was equal at each location, phase was probably the only condition.  You have put the poster on the right track.